Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
class ppq {
 public:
  int a[10];
  int n = 0;
  ppq() { fill(a, a + 10, INT_MAX); }
  void push(int x) { a[n++] = x; }
  int min1(int x) { return min(n * 1LL, x * 1LL); }
};
ppq merge(ppq one, ppq two) {
  int i = 0, j = 0, t = 0;
  ppq three;
  while (t < 10) {
    if (one.a[i] < two.a[j]) {
      three.a[t++] = one.a[i++];
    } else {
      if (one.a[i] == two.a[j]) i++;
      three.a[t++] = two.a[j++];
    }
    if (three.a[t - 1] == INT_MAX) {
      t--;
      break;
    }
  }
  three.n = t;
  return three;
}
int n, mm, q, a = 10;
struct node {
  vector<int> v;
  ppq id;
  vector<ppq> id1;
};
vector<node> adj(100005);
vector<node> up(100005);
vector<int> tin(100005), tout(100005);
int timer = 0;
using pq = ppq;
void dfs(int x, int p = 1) {
  tin[x] = timer++;
  up[x].v.resize(20);
  up[x].id1.resize(20);
  up[x].v[0] = p;
  up[x].id1[0] = adj[p].id;
  for (int i = 1; i < 20; i++) {
    up[x].v[i] = up[up[x].v[i - 1]].v[i - 1];
    up[x].id1[i] = merge(up[x].id1[i - 1], up[up[x].v[i - 1]].id1[i - 1]);
  }
  for (auto it : adj[x].v) {
    if (it != p) {
      dfs(it, x);
    }
  }
  tout[x] = timer++;
}
bool is_ancestor(int x, int y) {
  return tin[x] <= tin[y] && tout[x] >= tout[y];
}
pq lca(int u, int v) {
  pq ans = merge(adj[u].id, adj[v].id);
  for (int i = 19; i >= 0; i--) {
    if (!is_ancestor(up[u].v[i], v)) {
      ans = merge(ans, up[u].id1[i]);
      u = up[u].v[i];
    }
  }
  if (!is_ancestor(u, v)) {
    ans = merge(ans, up[u].id1[0]);
    u = up[u].v[0];
  }
  for (int i = 19; i >= 0; i--) {
    if (!is_ancestor(up[v].v[i], u)) {
      ans = merge(ans, up[v].id1[i]);
      v = up[v].v[i];
    }
  }
  return ans;
}
void print(pq x) {
  for (int i = 0; i < x.min1(a); i++) cout << x.a[i] << " ";
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> mm >> q;
  for (int i = 1; i < n; i++) {
    int v, u;
    cin >> v >> u;
    adj[v].v.push_back(u);
    adj[u].v.push_back(v);
  }
  for (int i = 1; i <= mm; i++) {
    int c;
    cin >> c;
    if (adj[c].id.n < 10) adj[c].id.push(i);
  }
  dfs(1);
  while (q--) {
    int v, u;
    cin >> v >> u >> a;
    pq aa = lca(v, u);
    cout << aa.min1(a) << " ";
    print(aa);
    cout << "\n";
  }
}